Explore Related Concepts

Best Results From Yahoo Answers Youtube


From Yahoo Answers

Question:tan x + sec x = 2 cos x solve for x (without a calculator) altho u guys can use a calculator, just the worksheet says not to i dont need a number, just the simplified form of the equation. thnx

Answers:tan x = sinx/cosx and sec x = 1/cosx so it becomes (sin x + 1)/cosx = 2 cosx. Cross multiply to get 2 cos^2 x = sinx + 1 Replace cos^2 x with its equivalent, 1 - sin^2 x to get 2(1 - sin^2 x) = sinx + 1. Distribute: 2 - 2 sin^2 x = sinx + 1. Move stuff from left to right: 0 = 2 sin^2 x + sinx - 1. Factor: 0 = (2 sinx - 1)(sin x + 1) So sin x = 1/2 or sin x = -1. Find where this happens and you have it.

Question:For Physics we have to do a trigonometry worksheet. I know how to do all of the problems except one. The triangle looks like this, http://i27.tinypic.com/uy71e.jpg we have to solve for, x, y and Q the triangle a right, 90 degree triangle. please help. thanks :)

Answers:I think you are missing something

Question:A square is a rectangle and a rhombus. Sometimes Always Never POINT VALUE: 1 points -------------------------------------------------------------------------------- If SQUA is a square, then Sometimes Always Never POINT VALUE: 1 points -------------------------------------------------------------------------------- A rhombus has opposite angles that are supplementary. Sometimes Always Never POINT VALUE: 1 points -------------------------------------------------------------------------------- A rectangle has diagonals that are congruent and perpendicular. Sometimes Always Never POINT VALUE: 1 points -------------------------------------------------------------------------------- If the diagonals of a rhombus are 10 cm and 24 cm, then the length of each side of the rhombus is 26 cm. Sometimes Always Never POINT VALUE: 1 points -------------------------------------------------------------------------------- In rhombus ABCD, the diagonals intersect at point E. If the measure of angle ABC = 84 degrees and AC = 18 cm, find the length of each side of the rhombus to the nearest tenth. (Hint: Draw and label a picture.) 12.1 cm 13.5 cm 18.1 cm 26.9 cm POINT VALUE: 3 points -------------------------------------------------------------------------------- ABCD is a rectangle whose diagonals intersect at point E. AC = 10, BC = x + 2, DC = x + 4. Find the perimeter of triangle DEC. 4 16 18 28 POINT VALUE: 3 points -------------------------------------------------------------------------------- In rectangle ABCD, AC = 20, BC = 12, DC = 3x y. and AD = 2x + y. Solve for x and y. Show all work to receive credit. WORTH 3 points (Not yet scored by your instructor) -------------------------------------------------------------------------------- Find the dimensions of rectangle RECT if the diagonals intersect at a 60 degree angle and RC = inches. Leave your answer in simplest radical form. Show work or explain your answer to receive credit. WORTH 3 points (Not yet scored by your instructor) -------------------------------------------------------------------------------- In square SQUA, the length of diagonal QA is 18 cm. In rhombus RHOM, the length of diagonal MH is 18 cm, and the measure of angle RHO = 60 degrees. Which figure has the larger perimeter? What is the difference between the perimeters? Show all work to receive credit. WORTH 3 points (Not yet scored by your instructor)

Answers:always (by definitions of square, rectangle and rhombus) --------------------------------------- sorry can't understand what you want --------------------------------------- sometimes (the opposite angles of a rhombus are equal thus only if it is a square) ---------------------------------------- sometimes (always congruent but perpendicular only if it is a square) ---------------------------------------- never (the diagonals of a rhombus bisect each other perpendicularly so for the diagonals, the side is 13cm) ---------------------------------------- you have AE = 9 cm (as the diagonals bisect each other perpendicularly) and angle ABE = 42 deg (the diagonals bisect the angles between two subsequent sides) thus the side = 9 / cos 42 = 12.1 -------------------------------------------- AC = 10 cm so BD = 10 cm (as the diagonals of a rectangle are congruent) so by Pythagoras theorem, 10^2 = (x + 2)^2 + (x + 4)^2, thus x = -3 + (89)^(1/2) or -3 - (89)^(1/2) which is rejected. thus the answer is EC + ED + CD = 5 + 5 + 1 + (89)^(1/2) = 11 + (89)^(1/2) i have triple checked my answer. check if you have typed the question correctly this is the method though. ------------------------------------------- AC = 20 BC = AD = 12 = 2x + y by Pythagoras theorem, you get 3x - y = 12 thus x = 4.8 and y = 2.4 ------------------------------------------ I'll tell you how to proceed as the value of RC is not typed. the subsequent angles at the point of intersection of the diagonals are supplementary. draw the perpendicular bisectors of the sides, they bisect the angles at the center X. let them intersect RE, EC, CT and RT at L, M, N and O now, RX = EX = CX = TX = RC/2 and angle RXO = 30 deg and angle RXL = 60 deg Use trigonometry to find RO (=RX cos30) and RL (=RX cos60). RT = 2 RO = EC RE = 2 RL = CT I will deal the same problem using the property of parallel lines (this is faster) after drawing the perpendicular bisectors, LN and MO you see that they are parallel to sides EC and CT respectively, and by corresponding angle property of parallel lines and transverse lines, angle RXO = angle RCT = 30 and angle RCE = 60 deg by complementary angle property of angles RCT and RCE again using trigonometry, RT = EC = RC cos30 and RE = CT = RC cos60 you may use pythagoras theorem in some places, but trigonometry is easier. ------------------------------------------------- the side of the square = diagonal / cos45 = 2^(1/2) * 18 thus the perimeter = 72 * 2^(1/2) = 101.8 as for the rhombus, draw the diagonal RO, they bisect perpendicularly at X. now angle RHX = 30 deg thus RH = HX / cos30 = 2 * 9 / 2^(1/2) = 6 * 3^(1/2) thus the perimeter = 24 * 3^(1/2) = 41.5 clearly the rhombus has the smaller perimeter.

Question:Last question on a hideously long worksheet...any help? The profit P for a product whose sales fluctuate with the seasons is estimated to be P=14+5sin( T/52), where T is given in weeks and P is in thousands of dollars. Determine the number of weeks it would take for the profit to initially reach $18,000. I keep getting stuck when I have the constant on one side and the sine on the other.

Answers:18 = 14 + 5sin( T/52) ----> T/52 = 0.9273 --->T = 15.3 weeks

From Youtube

Q #15 - June 2010 Algebra2 /Trigonometry Regents - pen and paper :This video shows the pen and paper method for this question from the June 2010 Algebra2/Trigonometry Regents. Visit www.tiskills.com for more videos and worksheets.

Q #9 - June 2010 Algebra2 /Trigonometry Regents - pen and paper :This video shows the pen and paper method for this question from the June 2010 Algebra2/Trigonometry Regents. Visit www.tiskills.com for more videos and worksheets.